= Solution
Differentiate the first <ordinary differential equation> and substitute the second:
$$
u''=au'+bv'=au'+bcu+bdv.
$$
The original first equation also gives $bv=u'-au$, so
$$
u''=(a+d)u'+(bc-ad)u.
$$
Therefore the requested <coefficients> are
$$
\boxed{\alpha=a+d,\qquad\beta=ad-bc}.
$$
They are the <trace> and <determinant> of the coefficient <matrix>. This derivation never divides by $b$ and is valid even if $b=0$. In that degenerate case the scalar second-order equation is a necessary consequence, while the original constraint $u'=au$ must also be retained when reconstructing the coupled system.
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